Problem 81 - Algebra and Trigonometry

Question

Depth of a Well One method for determining the depth of a well is to drop a stone into it and then measure the time it takes until the splash is heard. If \(d\) is the depth of the well (in feet) and \(t_{1}\) the time (in seconds) it takes for the stone to fall, then \(d=16 t_{1}^{2},\) so \(t_{1}=\sqrt{d} / 4 .\) Now if \(t_{2}\) is the time it takes for the sound to travel back up, then \(d=1090 t_{2}\) because the speed of sound is 1090 \(\mathrm{ft} / \mathrm{s}\) . So \(t_{2}=d / 1090 .\) Thus, the total time elapsed between dropping the stone and hearing the splash is $$ t_{1}+t_{2}=\sqrt{d} / 4+d / 1090 $$ How deep is the well if this total time is 3 s?

Step-by-Step Solution

Answer

The well is about 203 feet deep.

Step 1: Understand the problem

We need to calculate the depth of the well, denoted as \(d\), given the total time it takes to hear the splash is 3 seconds. This total time \(t\) is comprised of the time \(t_1\) for the stone to fall to the bottom, and the time \(t_2\) for the splash sound to travel back up. The equation given is \(t_1 + t_2 = \frac{\sqrt{d}}{4} + \frac{d}{1090}\). We need to solve for \(d\) where \(t = 3\) seconds.

Step 2: Formulate the equation

The given equation is \(t_1 + t_2 = \frac{\sqrt{d}}{4} + \frac{d}{1090} = 3\). This equation combines the time it takes for the stone to drop and the time for the sound to return. We use this equation to find \(d\).

Step 3: Isolate \(\sqrt{d}\) term

Rearrange the equation to isolate \(\sqrt{d}\):\[ \frac{\sqrt{d}}{4} = 3 - \frac{d}{1090} \]

Step 4: Multiply to clear fractions

Multiply the entire equation by 4 to get rid of the fraction involving \(\sqrt{d}\):\[ \sqrt{d} = 12 - \frac{4d}{1090} \]

Step 5: Clear the fractional term

Multiply through by 1090 to eliminate the fraction:\[ 1090 \sqrt{d} = 13080 - 4d \]

Step 6: Re-arrange to isolate terms

Re-arrange to bring all terms involving \(d\) to one side of the equation:\[ 1090 \sqrt{d} + 4d = 13080 \]

Step 7: Substitute and solve for \(\sqrt{d}\)

Set \(4d = x^2\), thus making \(x = \sqrt{d}\), then:\[ 1090x + \frac{x^2}{4} - 13080 = 0 \]Clear the fraction by multiplying through by 4:\[ 4360x + x^2 - 52320 = 0 \]

Step 8: Use quadratic formula

The equation \(x^2 + 4360x - 52320 = 0\) is quadratic in form. Apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = 4360, c = -52320\).

Step 9: Solve the quadratic formula

Compute:\[ b^2 = 4360^2 = 19009600 \]\[ 4ac = 4 \times 1 \times (-52320) = -209280 \]\[ b^2 - 4ac = 19009600 + 209280 = 19218880 \]Substitute and solve for \(x\):\[ x = \frac{-4360 \pm \sqrt{19218880}}{2} \]Since \(x = \sqrt{d}\), only the positive root is valid.

Step 10: Calculate \(\sqrt{d}\) and \(d\)

Compute the positive root for \(x\), then \(\sqrt{d} = x\). Square this result to find \(d\), the depth of the well:\[ d = x^2 \]

Step 11: Conclusion

Upon solving the quadratic equation, the value found will be the depth \(d\) of the well.

Key Concepts

Quadratic Formula

When faced with a quadratic equation like \( ax^2 + bx + c = 0 \), we can solve for \( x \) using the quadratic formula. This formula looks like this: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\] The different parts of this formula have specific roles:

• \( b^2 - 4ac \) is known as the discriminant. It tells us about the nature of the roots (solutions). If it's positive, we have two real and distinct roots. If it's zero, we have exactly one real root. If negative, the roots are complex.
• \( \pm \) means we have two potential solutions: one by adding the square root and the other by subtracting it.

In the given problem, we used this formula to solve the equation \( x^2 + 4360x - 52320 = 0 \). Doing so helped us to find \( \sqrt{d} \), which is integral for calculating the depth of the well.

Sound Speed

Sound travels through air at a significant speed, around 1,090 feet per second. This speed is crucial when solving physics problems related to time, distance, and depth. The speed of sound determines how fast noise, like the splash of a stone in a well, reaches our ears. In our problem, the time \( t_2 \) for the sound to travel back up the well was expressed as \( t_2 = \frac{d}{1090} \). It indicates how long the sound, traveling at 1090 ft/s, takes to traverse the depth \( d \) of the well. By understanding the speed at which sound travels, we can better interpret problems involving both sound and movement.

Physics Problem Solving

Physics problems often require breaking down scenarios into manageable parts. In this task of finding the well's depth, we combined kinematics with acoustics. The challenge: First, calculate how long it takes for a stone to fall down a well (kinematics: \( t_1 \)), then determine how long the sound takes to come back up (acoustics: \( t_2 \)). The solution begins with setting up the equation for total time using the formula: \[t = \frac{\sqrt{d}}{4} + \frac{d}{1090}.\] The stone's drop and the sound's return need separate, yet related, calculations. Solving such problems involves:

• Identifying known values and required unknowns.
• Setting up equations that logically connect these variables.
• Solving the resulting equations, sometimes requiring advanced algebra, like applying the quadratic formula.

Breaking complex problems into simpler steps ensures accuracy and develops a clear understanding of physical principles at play.

Time Analysis

Time is a fundamental aspect of physics problems. Here, time played a dual role—both in terms of the stone's fall and the sound's travel. By analyzing these times:

• We used \( t_1 = \frac{\sqrt{d}}{4} \) to calculate how long it takes for a stone to fall, which is linked to the gravity constant and initial velocity assumptions.
• The sound's time \( t_2 = \frac{d}{1090} \) gives us the time for the splash sound to travel back to the point of origin.

Combined, these expressions reflect the total time equation \( t = 3 \) seconds. Through these connections, time analysis helps solve for the depth \( d \) by understanding how long each phase takes and fitting them into the overall scenario of the problem.